Question

A 3.314−g sample of a mixture of cyclohexane (C6H12) and naphthalene (C10H8) is dissolved in 45.40 g of benzene (C6H6). The freezing point of the solution is 2.00 ° C. Calculate the mass percent of the mixture.

Answer 1

The molar masses of cyclohexane and naphthalene are 84 g/mol and 128 g/mol respectively.

In 3.314−g sample, let x g of a mixture of cyclohexane (C6H12) is present. 3.314-x g of naphthalene will be present.

The number of moles of cyclohexane $= \dfrac {x}{84}$

The number of moles of naphthalene $= \dfrac {3.314-x}{128}$

Total number of moles

$\\ \: \\ \: = \dfrac {x}{84} + \dfrac {3.314-x}{128} \\ \: \\ \: =\dfrac {128x+278.4-84x}{10752} \\ \: \\ \: =\dfrac {44x+278.4}{10752}$

45.40 g of benzene corresponds to 0.04540 kg.

The molality of solution is the number of moles of solute present in 1 kg of benzene.

Molality

$\\ \: \\ \: m=\dfrac {44x+278.4}{10752 \times 0.04540} \\ \: \\ \: m =\dfrac {44x+278.4}{488.14}$

The freezing point of the solution is 2.00 ° C.

The freezing point of benzene is 5.50 ° C.

The depression in the freezing point $\Delta T_f = 5.50-2.00=3.50 \: ^oC$

For benzene,

The depression in the freezing point

$\Delta T_f = K_f \times m$

$3.50 \: ^oC = 5.12 \: ^oC/m \times \dfrac {44x+278.4}{488.14}$

$x=1.257$

The mass percent of cyclohexane

$= 100 \times \dfrac {1.257}{3.314} \\ \: \\ \: =37.9$ %.

Hence, 37.9 % cyclohexane is present.

The mass percent of naphthalene